Whats my uncles age.

Puzzle:
I was visited my uncle.On my visit i met his friend as well.
I asked him what is their age.
My uncle replied saying,
the difference between their ages is 30 and if we multiply their ages it 1624.
What are their ages?


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Solution:
Lets assume the age of two uncles is x and y
based on the conditions

x-y=30
xy=1624 --->y=1624/x
by substituting second equation into the first one

x-1624/x=30

multuply by x on  both sides

x2 -1624=30x
x2-30x-1624=0
this is a quadratic equation.like ax2+bx+c=0 the roots can be found by using the formula
you can follow the link on how to solve the equation http://www.wikihow.com/Solve-Quadratic-Equations

only positive root of the equation is 58.

so tha ages are 58 and 28
 Corrected My Answer:






Winner:Nithya is the Winner

Split them in half.

Puzzle:
A table full of quarter coins(infinite) are placed in front you.out of them 10 coins are showing tails and remaining are showing heads.
You are blind folded and you are wearing gloves.so you can not see and feel which coins are heads and tails.

Split the Quarters into two piles so that each pile has exactly same number of tails?


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Solution: The trick here is the piles does not have to be off the same size.Take exactly 10 coins from the whole pile and make it as second pile.In the second pile flip each coin once.
Now the two piles will have same number of trails.
Winner:

Zebra Puzzle.

Puzzle:

1. There are five houses.
2. The Englishman lives in the red house.
3. The Spaniard owns the dog.
4. Coffee is drunk in the green house.
5. The Ukrainian drinks tea.
6. The green house is immediately to the right of the ivory house.
7. The Old Gold smoker owns snails.
8. Kools are smoked in the yellow house.
9. Milk is drunk in the middle house.
10. The Norwegian lives in the first house.
11. The man who smokes Chesterfields lives in the house next to the man with the fox.
12. Kools are smoked in the house next to the house where the horse is kept.
13. The Lucky Strike smoker drinks orange juice.
14. The Japanese smokes Parliaments.
15. The Norwegian lives next to the blue house.

Now, who drinks water? Who owns the zebra?

In the interest of clarity, it must be added that each of the five houses is painted a different color, and their inhabitants are of different  national extractions, own different pets, drink different beverages and smoke different brands of American cigarets [sic]. One other thing: in statement 6, right means your right.

Courtesy:Wiki Life International, December 17, 1962

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Solution:

House	5	4	3	2	1
Color	Ivory	Green	Red	Blue	Yellow
Nationality	Spaniard	Japanese	Englishman	Ukrainian	Norwegian
Drink	Orange juice	Coffee	Milk	Tea	Water
Smoke	Lucky Strike	Parliament	Old Gold	Chesterfield	Kools
Pet	Dog	Zebra	Snails	Horse	Fox

Step 1

We are told the Norwegian lives in the first house (10). We count from left to right.

From (10) and (15), the second house is blue. What color is the first house? Not green or ivory, because they must be next to each other (6 and the second house is blue). Not red, because the Englishman lives there (2). Therefore the first house is yellow.

It follows that Kools are smoked in the first house (8) and the Horse is kept in the second house (12).

So what is drunk by the Norwegian in the first, yellow, Kools-filled house? Not tea since the Ukrainian drinks that (5). Not coffee since that is drunk in the green house (4). Not milk since that is drunk in the third house (9). Not orange juice since the drinker of orange juice smokes Lucky Strikes (13). Therefore it is water (the missing beverage) that is drunk by the Norwegian.

Step 2

Who can live in the second, the blue, house? First, we note that the Japanese cannot drink water (as we deduced the Norwegian drinks this), nor tea (5) nor orange juice (13), as he does not smoke Lucky Strike. So, the Japanese drinks either coffee (4) or milk (9), which also means he is either in the green house or in the middle house - and we know the second house (the blue one) is neither green nor in the middle. So, the Japanese cannot reside in the blue house, which leaves us only with the Ukrainian.

So what does the Ukrainian smoke? It cannot be Kool (8 taken by Norwegian), nor Old Gold (7 we now know the Ukrainian owns the horse), nor Lucky Strike (13, 5 the Ukrainian drinks tea), nor Parliaments (14). Thus he smokes Chesterfields.

Step 3

We can now assign the remaining 2 cigarette brands to the respective nationality: Since Old Gold is smoked by the snails owner (7), the Spaniard cannot smoke the brand (3 he owns a dog). Thus the English smokes Old Gold (and owns snails) and the Spaniard smokes Lucky Strike.

Step 4

We deduced in the previous step that the Spaniard smokes Lucky Strike, which implies he drinks orange juice (13). With this, the only unassigned drinks left are coffee and milk. Since the Englishman lives in the red house (2), he cannot be the one drinking coffee in the green house (4). Thus, the Englishman drinks milk in the middle house (9), which is red (2). And so, the Japanese drinks coffee in the green house (4).

Now we assigned all the drinks, and we also can assign the missing houses. We just concluded that the Englishman lives in the middle, red house and the Japanese in the green house. The ivory house is the only one left unassigned, and it must be the Spaniard's. Furthermore, since the green house is to the right of the ivory house (6) it is the last (i.e. rightmost) house, and we now have the order of houses (from left to right): yellow, blue, red, ivory, green.

Step 5

What is left to do is to assign the remaining 2 animals (fox, zebra) to their respective owners (Norwegian, Japanese) (we deduced in step 2 that the Ukrainian owns the horse, and instep 3 that the Englishman owns snails, and we already know that the Spaniard owns a dog (3)).

We know that the "man who smokes Chesterfields lives in the house next to the man with the fox" (11). The Ukrainian smokes Chesterfields and is neighbors with the Norwegian, but not with the Japanese. Thus, the Norwegian owns the fox and the Japanese owns the zebra.

Winner:

Give me the gold

Puzzle:
We have 5 bags of coins.each bag has 50 coins.only one bag has gold coins in it other bags has fake gold coins.each fake coin weighs 1 gram,.each real coins weighs 1.01 grams.
we have a scale and we can use it only once.how to identify which bag has real gold coins.


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Solution:place the following coins in scale
1 coin from bag 1
2 coins from bag 2
3 coins from bag 3
4 coins from bag 4 and 
5 coins from bag 5 
For example, if the 5 coins weigh 15.05 grams, there must be 5 gold coins. These will have been from bag number 5 if coins weigh 15.04 the gold coins in 4th bag Winner:Aswani

Measure the time.

Puzzle:
You have given two ropes of different lengths and a lighter.
If you light the rope at one end,each rope takes 1 hours to burn completely.
though it takes same amount of time to burn the ropes they burn at different speed.
Meaning, first half of the first rope may take 15 minutes to burn and the second half may take 45 minutes to burn.
and this rate is different for the second rope.

Now the question is how do you measure 45 minutes of time using the two ropes and the lighter?

Solution:Burn the first rope from both sides and burn the second rope from one side.
First rope burns completely in 30 minutes as we lit it from both ends.
as soon as the first rope completely burned lit the other end of the second rope.it will take 15 minutes more to completely burn the second rope. 30+15 is your 45 minutes.

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Solution:
Winner:

Which bulb is not working.

Puzzle:
You are standing in the first floor of the building next to 3 light switches.There is one room in the second floor of the building, where there are 3 light bulbs.Each light bulb is operated by the one of the switches next to you.Because the light bulbs are in another room you can not see them,

Now the question: One of the bulbs is not working.You can only turn on one switch and you can only make one trip to the room where the bulbs are.How to find out which bulb is not working.

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Solution: Turn one switch on.
Case 1: If a bulb is on then the power is flowing to that bulb socket.take that bulb out and try the remaining two bulbs in the working socket.the bulb that is not working is the faulty one
Case 2:If no bulb is glowing.Take all three bulbs out.Try each bulb in 3 sockets,faulty bulb will not work in any of the socket.

Winner:

Let me enter

Puzzle:

A man is walking on the road.he saw a cool party going on.what he saw is everyone trying to enter the party is  given a set of numbers and the persons giving the correct answers are allowed to the party.

He overheard couple of numbers and correct answers
Numbers     Answers
8903   --- 4
1100   --- 2
2102   --- 1
7211   --- 0
0836   --- 4
8898   --- 7

He thinks he heard enough.He went to attend the party.
His number is 9822.He replied and he entered the party.What was his answer.


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Solution: answer is 3 .Logic is Number of closed loops in the numbers given.
Winner:

Distance covered - One puzzle a day - Puzzle Buddies

Puzzle:There is a 50 m long army platoon marching ahead. The last person in the platoon wants to give a letter to first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved by 50 m.

The question is how much distance did the last person cover in that time assuming that he ran the whole distance with uniform speed.
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Solution:The last person covered 120.71 meters.

It is given that the platoon and the last person moved with  uniform speed. Also, they both moved for the identical  amount of time. Hence, the ratio of the distance they  covered - while person moving forward and backword - are  equal.   Let's assume that when the last person reached the first  person, the platoon moved X meters forward.   Thus, while moving forward the last person moved (50+X)  meters whereas the platoon moved X meters.   Similarly, while moving back the last person moved [50-(50- X)] X meters whereas the platoon moved (50-X) meters.   Now, as the ratios are equal,  (50+X)/X = X/(50-X)  (50+X)*(50-X) = X*X   Solving, X=35.355 meters   Thus, total distance covered by the last person  = (50+X) + X  = 2*X + 50  = 2*(35.355) + 50  = 120.71 meters  

Winner:

Coin Tossing: One puzzle a day - Puzzle Buddies

Puzzle: There's a coin toss game in a casino. You are friends with the dealer and can set up a deal with him. Dealer can get what he wants 80% of times. The bet is that if you win you win $1 and if you loose then you loose $1. You can make X attempts only. What should be your strategy to always win.

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Source: CSE Blog

Camel and Bananas - One puzzle a day - Puzzle Buddies

Puzzle:

A banana plantation is located next to a desert. The plantation owner has 3000 bananas which he wants to transport to the market by camel. The market is located 1000 kilometre away from the plantation with desert in between. The owner has only one camel, which carries a maximum of 1000 bananas at any moment in time. The camel eats one banana every kilometre it travels.

What is the largest number of bananas that can be delivered at the market?

Answer:

The camel needs to make more than one trip to deliver all bananas to the market. No doubt he needs to make at least one stop in between otherwise he can never have enough bananas left for the comeback journey. And this point needs to be less than 500 if he wants any bananas to be left when he reaches the market. Let’s say it’s A.

Plantation======A=====================Market

The trick is to make sure that the camel always travels with his full capacity of bananas. This is true initially as there are total 3000 bananas and the camel carries 1000 bananas at a time. So, he will make 3 trips with 1000 bananas to point A. To make sure he carried his full capacity after point A, we should have a multiple of 1000 bananas left when all bananas are at point A. This number can thus be 2000 at best. So he is eating 1000 bananas from the plantation to point A, making 5 trips in between. So, he eats 200 bananas in 1 trip. Point A is 200 km from the plantation.

Again we need a stopping point in between A and the plantation where there will be 1000 bananas left after 3 trips. So the camel eats 333 1/3 bananas in each trip, point B being 333 1/3 km away from A.

Plantation======A=========B===========Market

In the end, the camel needs to make only 1 trip from B to the market which is 466 2/3 km away, leaving 533 1/3 bananas at the market.

Winner: Ashish and Prabharan

How many Days- One Puzzle a Day -Puzzle Buddies

Puzzle: There is a frog lying under the bottom of a well of depth 40 meters.every day morning the frog leaps up 4 meters and it sleeps till next morning.while the frog is asleep he falls 3 meters down.how many days it will take the frog to escape from the well?

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Solution: on first day the frog jumps from 0 t0 4 meters and by night it slips 3 meters

so day 2 it starts at 1

day 3 it starts at 2 likewise

day 37 it starts at 36 and it jumps 4 meters and it comes out

So it takes 37 days.

Winner: Gaurave

Pair game - One puzzle a day - Puzzle Buddies

Puzzle:Ganesh has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Ganesh likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Ganesh finds at least one pair of cards that have the same value?
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Solution:The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1.

First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt).

Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. We have 11 cards left in the deck.

Probability of NO pairs so far = 10/11.

Third card: Since we have gotten no pairs so far, we have two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. We have 10 cards left in the deck.
Probability of turning over a third card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 8/10.

Cumulative probability of avoiding a pair BOTH on the second card AND on the third card = product of the two probabilities above = (10/11) (8/10) = 8/11.

Fourth card: Now we have three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. We have 9 cards left in the deck.
Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 6/9.

Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.

Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 - 16/33 = 17/33.
Winner: Nonone

How many weights? Part3 - One puzzle a day - Puzzle Buddies

Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.

You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.

The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.

Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.

Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.

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Source: Classic Puzzles

Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1

To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2

Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.

To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4

Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.

Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.


Winner: Kasturi again is the winner.

Boat and Anchor - One Puzzle a Day - Puzzle Buddies

Puzzle:

A fisherman is sitting in his boat in the center of a small lake. He throws his anchor into the water. Does the water level rise, fall or stay the same?

Solution:

When the anchor is in the boat, the water displaced is equal to the weight of the anchor. When the anchor is dropped in the water, the water displaced is equal to the volume of the anchor. Since the anchor is denser than water, its weight is more than its volume as compared to water. Hence, it displaces less water when it is thrown in the lake. Thus, the water level falls.

Winner: Vinay

Puzzle: Palindrome - One Puzzle a Day -Puzzle Buddies

Puzzle:

What is smallest possible palindrome number whose cube root is not a palindrome?

For example 1331 is a palindrome. Cube root of 1331 is 11 which is also a palindrome.so this does not satisfy the condition.

Note:a palindrome is a word,phrase or a number that can be read same in either direction.

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Solution:10662526601 = 2201³.It is not yet known that if there are any other palindromic numbers whose cube roots are not palindromes.

As of now there are no palindromes of the form X ⁿ where X is not a palindrome and n>3.

the only exception is 2201.

Winner:

Passenger seating - One puzzle a day - Puzzle Buddies

Puzzle:A line of 100 airline passengers is waiting to board the plain. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let's say that the Nth passenger in line has a ticket for the seat number N.)
Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
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Solution:The solution to this one is pretty mathimatically involved.

The crazy passenger that boards first has an equal chance of getting on any of the 100 seats. Therefore, the probability of this person sitting in a particular seat k is 1/100.

Now for any value of k, all the passengers up to k will board and take their seats as normal. Then the k-th passenger will find themselves out of a seat, and pick a random seat, in effect, becoming the crazy passenger for a problem with 101 - kpassengers.

(For example if the crazy person takes the tenth seat, passengers 2 through 9 will be seated normally, and when passenger 10's turn comes, they can be considered the crazy passenger first in line of 91 people boarding a plane with 91 seats.)

Let us define P(n) as the probability that the last of n passengers boards their proper place under the conditions of the problem. So for solution, we are looking to find the value of P(100).

In a situation with n passengers, the first, "crazy", passenger can take any of the seats 1 .. n with the probability of 1/n. Also, given some seat number k that the crazy passenger may occupy, the problem will become that of 1+n-k passengers. Therefore, we can say that

P(n) = 1/n * (1 + P(n-1) + P(n-2) + ... + P(2))

(This problem does not make sense with less than 2 passengers. Also, it is easy to see that P(2) is 1/2)

What we would like to do is see what the value of P(n) is in terms of P(n-1).

P(n-1) = 1/(n-1) * (1 + P(n-2) + P(n-3) + ... + P(2) P(n-1) * (n-1) = (1 + P(n-2) + P(n-3) + ... + P(2) P(n) = 1/n * (1 + P(n-1) + P(n-2) + ... + P(2)) (P(n) * n) - P(n-1) = (1 + P(n-2) + ... + P(2))

So,

(P(n) * n) - P(n-1) = P(n-1) * (n-1) (P(n) * n) - P(n-1) = (P(n-1) * n) - P(n-1) (P(n) * n) = (P(n-1) * n) P(n) = P(n-1)

This shows us that the problem's answer does not change if we add one more passenger. Starting at 2 passengers with the anser of 1/2, we can add passengers indefinitely, and still arrive at the same answer. Thus for the instance with 100 passengers, the probability of the last one getting to their proper seat is also 1/2, or 50%.
Winner:

Underground Cable - One Puzzle a Day - Puzzle Buddies

Puzzle:

There is an underground cable in which there are 10 wires, no two wires being connected in any way. The two ends of the cable are 5 miles apart. Sad thing is the wires are all of the same color, material and width and are not labeled. You cannot figure out which wire at one end corresponds to which wire at the other end by looking at them.

You have been given the job to label all the wires at both the ends. Armed with only a battery and a bulb and assuming that you are standing at one of the ends, how many total miles will you have to walk to finish your job? (Don't ask why you don't have a car!)

Will the number change if there are 20 wires?

Solution:

Call the end of the cable that you are at as X. Label one wire A. Join two wires together and label both of them B. Join three wires together and label each of them C. Join the remaining four wires together and label each of them D.

Now go to the other end Y. With the battery and the bulb you can easily find out which wires are joined together and label them as A, B, C or D accordingly.

Now join together one wire from each of the groups and label each of them as 1. Join one wire from each of the groups B, C and D and label them as 2. Join one wire each from C and D and label them as 3. Label the remaining wire from D as 4. We now have the following labels at Y end of the cable: A1, B1, C1, D1, B2, C2, D2, C3, D3 and D4.

Now go to the other end X. Find out the wires joined together and label them as 1, 2, 3 or 4 respectively. We had previously labeled wires as A, B, C or D at end X. So, we now have the labels A1, B1, C1, D1, B2, C2, D2, C3, D3 and D4 at end X too.

It is easy to verify that similarly labeled wires at the two ends are one and the same.

Thus, you had to make only one trip, walking 10 miles. This logic can be applied to any number of wires and the walking distance remains the same.

How many weights? Part2 - One puzzle a day - Puzzle Buddies

Puzzle: This is part 2 of puzzle posted yesterday. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.
You can check Part 1 of this puzzle How many weights? Puzzle Part 1.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.

Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.
We can start from smallest number 1. To measure 1 we definitely need 1.
To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.
Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.

Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.
Winner: Kasturi wins again with her excellent explanation.

Eight Queens - One puzzle a day - Puzzle Buddies

Puzzle: How many ways we can arrange eight queens on a 8*8 chessboard, so that no queen can kill the other queens?

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Solution:puzzle has 92 distinct solutions. If solutions that differ only by rotations and reflections of the board are counted as one, the puzzle has 12 solutions

How many weights? - One puzzle a day - Puzzle Buddies

Puzzle: You have a common balance with two sides. You can only keep weighs on one side. You have to weigh  items from 1 to 1000 weight units with integral weights only.
What is the minimum number of weights required and what will be their weights?
Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.
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Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.
So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.
So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.
Winner: Kasturi is again the winner.